University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 65


$$\frac{dy}{dt}=-t\sin t^2(1+\cos(t^2))^{-1/2}$$

Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}\sqrt{1+\cos(t^2)}=\frac{d}{dt}(1+\cos(t^2))^{1/2}$$ Following the Chain Rule: $$\frac{dy}{dt}=\frac{1}{2}(1+\cos(t^2))^{-1/2}\frac{d}{dt}(1+\cos(t^2))$$ $$\frac{dy}{dt}=\frac{1}{2}(1+\cos(t^2))^{-1/2}(0-\sin t^2\frac{d}{dt}(t^2))$$ $$\frac{dy}{dt}=\frac{1}{2}(1+\cos(t^2))^{-1/2}(-2t\sin t^2)$$ $$\frac{dy}{dt}=-t\sin t^2(1+\cos(t^2))^{-1/2}$$
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